Question

If x =√3+√2/√3-√2 and y =1 ,the value of x-y/x-3y is?

Answer 1

x = √3+√2 / √3 -√2

Rationalizing the denominator

$x = \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} }$

$x = \frac{( { \sqrt{3} + \sqrt{2} ) {}^{2} } }{( \sqrt{3} ) {}^{2} - ( \sqrt{2} ) {}^{2} }$

$x = \frac{3 + 2 + 2 \times \sqrt{3} \times \sqrt{2} }{3 - 2}$

$x = \frac{5 + 2 \sqrt{6} }{1}$

$\boxed{ \boxed{x = 5 + 2 \sqrt{6} }}$

y = 1

$\frac{x - y}{x - 3y}$

$= \frac{5 + 2 \sqrt{6} - 1}{5 + 2 \sqrt{6} - 3 \times 1 }$

$= \frac{4 + 2 \sqrt{6} }{5 + 2 \sqrt{6} - 3}$

$= \frac{4 + 2 \sqrt{6} }{2 + 2 \sqrt{6} }$

$= \frac{2(2 + \sqrt{6}) }{2(1 + \sqrt{6} )}$

$= \frac{2 + \sqrt{6} }{1 + \sqrt{6} }$

Rationalizing the denominator again

=[2+√6/1+√6 ][1-√6/ 1-√6]

= [(2+√6)(1-√6)]/[(1+√6)(1-√6)]

= [2-2√6 +√6 - 6]/[1-6]

= [ -4-√6]/[-5]

=-[4+√6]/[-5]

= 4+√6 /5

Answer 2

$\huge \underline{ \rm \red{hello}}$

Question:-

If x =√3+√2/√3-√2 and y =1 ,the value of x-y/x-3y is?

Solution:-

x = (√3 + √2) / (√3 - √2)

= {(√3 + √2)(√3 + √2)} / {(√3 + √2)(√3 - √2)}

= (3 + 2 + 2√6) / (3 - 2)

= 5 + 2√6

y = 1

Now, (x - y) / (x - 3y)

= (5 + 2√6 - 1) / (5 + 2√6 - 3)

= (4 + 2√6) / (2 + 2√6)

= 2(2 + √6) / 2(1 + √6)

= (2 + √6) / (1 + √6)

= {(2 + √6)(1 - √6)} / {(1 + √6)(1 - √6)}

= (2 - √6 - 6) / (1 - 6)

= (- 4 - √6) / (- 5)

= (4 + √6) / 5